A community for students.
Here's the question you clicked on:
 0 viewing
aroub
 3 years ago
Why do we have to put an absolute value in the final answer?!
aroub
 3 years ago
Why do we have to put an absolute value in the final answer?!

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0final answer to what?

aroub
 3 years ago
Best ResponseYou've already chosen the best response.1\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can't even see that, is it 10 th root?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0see moderator has warned me now...whats the way to delet ur account...can somebody give the link...i m going frm here

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0can u pls give me the link...i want to delet my account..i can't find it anywhere

aroub
 3 years ago
Best ResponseYou've already chosen the best response.1I don't know.. You have to contact a moderator

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0cmon moderator show up now...

phi
 3 years ago
Best ResponseYou've already chosen the best response.0The only reason I can think of is they want to make sure you get a real and postiive (as opposed to complex) root. There will be 10 different roots.

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1@akash_809 you cannot delete OS accounts

aroub
 3 years ago
Best ResponseYou've already chosen the best response.1But you can write it without absolute value, right? Because (or this is what I've been taught) there's x^3 in the radicand so it's definitely positive..

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1the simple reason is that\[x^{10}=(x)^{10}\]so \[\large\sqrt[10]{x^{10}}\]will give you the same answer whether x is negative or positive

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what...thats cynical...open study community is surely going nowher with all these warnings and suspensions...and to top that off , you even cant delete your account...what is this open study team please response...i wnat to know how can i delete my account

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1absolute value of x is actually defined in 2 ways: 1) distance from the point x=0 on the number line 2) \(x=\sqrt{x^2}\)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1so since\[\sqrt{(x)^2}=\sqrt{x^2}=x\]then\[\sqrt[10]{x^{10}}=\sqrt[10]{(x)^{10}}=x\](because x to any even power is positive, whether x>0 or x<0)

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1does that help you @aroub ?

aroub
 3 years ago
Best ResponseYou've already chosen the best response.1In this case I totally understand why. But in the question itself there's an odd exponent in the radicand so it's definitely positive right? So then no need for an absolute value..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@TuringTest root of negative number is not defined...it will result in iota value...so x wont be the answer i think

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1not necessarily, let's assume x<0...

aroub
 3 years ago
Best ResponseYou've already chosen the best response.1If you assumed that, the question will be wrong.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes it is...\[lets take an example......\sqrt{2} is \not defined unless you are taking imaginary numbers into account then u wl get +1.414 i\]

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1the root of a negative number is not undefined, it is imaginary...\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sry reply got messed up i was just saying that square root of 2 is not defined unless you consider imaginary numbers as wel..

aroub
 3 years ago
Best ResponseYou've already chosen the best response.1Put imaginary numbers on a side for now..

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1imaginary numbers are defined... now if x<0, then x^3<0 that means that (x^3)(x^3)>0 which is what we would expect given that x^6>0 since all numbers to a positive exponent are >0

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1another way to see it, let's assume x>0\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}=\sqrt[10]{x^{10}}=x\]now we do it with the negative of x\[\sqrt[10]{(x)^4} \times \sqrt[10]{(x)^3} \times \sqrt[10]{(x)^3}=\sqrt[10]{(x)^4}\times\sqrt[10]{(x)^3\cdot (x)^3}\]\[=\sqrt[10]{(x)^{10}}=\sqrt[10]{x^{10}}=x\]so you see, after all is said and done we do not know whether the x we had at the beginning was positive or negative, hence we can only conclude that the answer is x

aroub
 3 years ago
Best ResponseYou've already chosen the best response.1Okay take this as another example: \[4\sqrt[4]{5a^2}\times 8\sqrt[4]{125a}\times 2\sqrt[4]{a}\] Here, a does not need an absolute value. Why? You can assume that a>0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yes you would have to assume a >0

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1let's just look at powers of a...

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1if a>0\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=a=a\](since we already knew a>0 now with the negative of a...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea i m waiting for that negative values of a...what would it give

aroub
 3 years ago
Best ResponseYou've already chosen the best response.1Aoh, so you have to try the negative and the positive?

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1\[\sqrt[4]{(a)^2}\times\sqrt[4]{a}\times\sqrt[4]{a}=\sqrt[4]{a^4}=a\neq a\]because a>0, but we just assumed a<0 ! a contradiction, hence the assumption makes a difference

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1sorry, I mean we took the negative of a...

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1let me write that again, gotta be careful here

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's the problem...the thing you did here is called fallacy of maths... http://en.wikipedia.org/wiki/Mathematical_fallacy ....search invalid roots in this article

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1assume a>0\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=a=a\]assume a<0, then\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=a\neq a\]since we assumed a<0

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1@akash_809 yes but at least one of these numbers is positive because it is raised to an even power, so I have committed no fallacy

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0did u went through the link......the proofs given like \[1=\sqrt{1}=\sqrt{(1)(1)}=\sqrt{1}\sqrt{1}=i.i=1 ....so u hv 1=1...and certainly \it is \not valid..\]

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1same thing with the earlier problem...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and 1=1....surely not right...

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1yes, but\[\sqrt{(1)^2}\cdot\sqrt{1}=\sqrt1\cdot\sqrt{1}\] is not the same as\[\sqrt{1}\cdot\sqrt{1}\]

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1notice we have a^2 in the above, so we know *that* part is positive

TuringTest
 3 years ago
Best ResponseYou've already chosen the best response.1this turned out to be more subtle than I bet you realized @aroub :)

aroub
 3 years ago
Best ResponseYou've already chosen the best response.1lol, yeah true xD ALWAYS thought absolute values are so easy! And whenever someone asks me about absolute values I answer until this question came up .' This question is a bit confusing >.< Well, THANKS LOOAADSSSSSS :D
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.